A round flat metal coil has 180 turns and negligible resistance.It is connected in a series circuit with a $17 - \Omega$ resistor,with nothing else in the circuit.You measure that a $6.0 - \mathrm { A }$ current flows through the resistor when a magnetic field through the coil,perpendicular to its area,is changing at $3.0 \mathrm {~T} / \mathrm { s } .$ What is the radius of the coil?

Question

Quizplus · Accepted Answer

The emf in the coil is given by:$$\varepsilon = -N\frac{d\Phi_B}{dt} = -N\pi r^2\frac{dB}{dt}$$where $N$ is the number of turns, $r$ is the radius of the coil, and $B$ is the magnetic field. The current in the circuit is given by:$$I = \frac{\varepsilon}{R} = \frac{N\pi r^2}{R}\frac{dB}{dt}$$Solving for $r$, we get:$$r = \sqrt{\frac{IR}{N\pi\frac{dB}{dt}}}$$Substituting the given values, we get:$$r = \sqrt{\frac{(6.0 \mathrm{ A})(17 \Omega)}{(180)(3.14)(3.0 \mathrm{ T} / \mathrm{ s})}} = 0.25 \mathrm{ m}$$Therefore, the correct choice is A.

A Round Flat Metal Coil Has 180 Turns and Negligible 17−Ω17 - \Omega17−Ω

A Round Flat Metal Coil Has 180 Turns and Negligible $17 - \Omega$